题目

Problem

The curve with equation $y=f(x)$ , $x>0$ , passes through the point $P(4,-2)$ .

Given that

\begin{align*} \frac{dy}{dx}=3x\sqrt{x}-10x^{-1/2}, \end{align*}

(a) find the equation of the tangent to the curve at $P$ , writing your answer in the form $y=mx+c$ , where $m$ and $c$ are integers to be found.

(4)

(b) Find $f(x)$ .

(5)

解答

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切线斜率是 $\frac{dy}{dx}$ 在 $x=4$ 时的值。再用点 $P(4,-2)$ 写直线。

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At $x=4$ ,

\begin{align*} \frac{dy}{dx} =&\,3(4)\sqrt4-10(4)^{-1/2}\\ =&\,24-5\\ =&\,19. \end{align*}

\begin{align*} y+2=&\,19(x-4)\\ y=&\,19x-78. \end{align*}

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把导函数写成指数形式后积分，再用 $P(4,-2)$ 求常数。

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\begin{align*} \frac{dy}{dx}=3x^{3/2}-10x^{-1/2}. \end{align*}

Integrating,

\begin{align*} f(x) =&\,3\cdot\frac{x^{5/2}}{5/2} -10\cdot\frac{x^{1/2}}{1/2} +c\\ =&\,\frac65x^{5/2}-20x^{1/2}+c. \end{align*}

Use $f(4)=-2$ :

\begin{align*} -2 =&\,\frac65(4)^{5/2}-20(4)^{1/2}+c\\ =&\,\frac{192}{5}-40+c\\ =&\,-\frac85+c. \end{align*}

Hence

\begin{align*} c=-\frac25. \end{align*}

Therefore

\begin{align*} f(x)=\frac65x^{5/2}-20\sqrt{x}-\frac25. \end{align*}